Find the net area and the area of the region bounded by \( y=10 \cos x \) and the \( x \)-axis between \( x=-\frac{\pi}{2} \) and \( x=\pi \). The net area is \( \square .( \) Simplify your answer.)
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To find the net area and the area of the region bounded by \( y=10 \cos x \) and the \( x \)-axis between \( x=-\frac{\pi}{2} \) and \( x=\pi \), we first need to identify the behavior of the function within this interval. 1. **Finding the intersections**: - \( y = 10 \cos x \) intersects the \( x \)-axis where \( 10 \cos x = 0 \), or \( \cos x = 0 \). - This occurs at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \) within the given interval. 2. **Calculating the net area**: - The area above the \( x \)-axis is between \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \), and the area below the \( x \)-axis is between \( x = \frac{\pi}{2} \) and \( x = \pi \). The function is positive from \( x = -\frac{\pi}{2} \) to \( x = \frac{\pi}{2} \) and negative from \( x = \frac{\pi}{2} \) to \( x = \pi \). 3. **Calculating the areas**: - Area from \( x = -\frac{\pi}{2} \) to \( x = \frac{\pi}{2} \): \[ A_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 10 \cos x \, dx = 10 [\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 10 [\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})] = 10 [1 - (-1)] = 20 \] - Area from \( x = \frac{\pi}{2} \) to \( x = \pi \): \[ A_2 = \int_{\frac{\pi}{2}}^{\pi} 10 \cos x \, dx = 10 [\sin x]_{\frac{\pi}{2}}^{\pi} = 10 [\sin(\pi) - \sin(\frac{\pi}{2})] = 10 [0 - 1] = -10 \] We take the absolute value since we want the area: \[ A_2 = 10 \] 4. **Net area**: \[ \text{Net Area} = A_1 + A_2 = 20 - 10 = 10 \] Thus, the net area is \( \boxed{10} \). The total area bounded by the curve and the \( x \)-axis is \( 20 + 10 = 30 \).
